Anyone who writes LaTeX documents using the AMS-LaTeX packages
(amsmath, amsthm, amssymb) and wants to convert these
documents to Markdown format to use with MathJaX. These Markdown files
can then be easily added to any web platform - Jekyll blogs, Wordpress,
basic HTML sites, etc.

In short, if you seek to use MathJaX to view your LaTeX documents
online, then you might be interested in this.

For a working example, have a look at the source of the
tullo.ch homepage
here.

Why not use Pandoc?

Pandoc
is an excellent document converter for less complex LaTeX documents.
Indeed, I’ve used it to convert this README document to a reST version
for use on PyPI.

Unfortunately, it is not designed to deal with documents that use the
AMSTeX extensions - which include the theorem, lemma, proof, and
exercise environments that are heavily used for typesetting papers,
lecture notes, and other documents.

As neither Pandoc nor MathJaX can deal with these documents, I hacked
together a set of regular expressions that can convert a subset of LaTeX
to Markdown, and used a few more to convert the sMarkdown to
MathJaX-convertible Markdown.

Example

As an example, the following LaTeX code:

\section{Example Section}
\begin{thm}[Euclid]
There are infinitely many primes.
\end{thm}
\begin{proof}
Suppose that $p_1 < p_2 < \dots < p_n$ are all of the primes.
Let $P = 1 + \prod_{i=1}^n p_i$ and let $p$ be a prime dividing $P$.
Then $p$ can not be any of $p_i$, for otherwise $p$ would divide the
difference $P - \left(\prod_{i=1}^n p_i \right) - 1$, which is impossible.
So this prime $p$ is still another prime, and $p_1, p_2, \dots p_n$
cannot be all of the primes.
\end{proof}

is converted into the following Markdown:

### Example Section
#### Theorem 1 (Euclid)
> There are infinitely many primes.
#### Proof
Suppose that $p_1 < p_2 < \dots < p_n$ are all of the primes.
Let $P = 1 + \prod_{i=1}^n p_i$ and let $p$ be a prime dividing $P$.
Then $p$ can not be any of $p_i$, for otherwise $p$ would divide the difference
$P - \left(\prod_{i=1}^n p_i \right) - 1$, which is impossible. So this prime
$p$ is still another prime, and $p_1, p_2, \dots p_n$ cannot be all of the primes.